Defining Inner Products

This questions has been stumping me for a few days now. But the questions asks how I could define an inner product for a vector space V such that all the vectors in a basis C are orthonormal.

So all vectors in a basis will have inner products equal to 0 (orthogonal), but how can I define another inner product to make sure that their length is 1, without dividing the inner product by the inner product..

Re: Defining Inner Products

This questions has been stumping me for a few days now. But the questions asks how I could define an inner product for a vector space V such that all the vectors in a basis C are orthonormal.

So all vectors in a basis will have inner products equal to 0 (orthogonal), but how can I define another inner product to make sure that their length is 1, without dividing the inner product by the inner product..

I can't really visualize this. Some insight would be appreciated.

As evry vector can be writen (uniquely) as a linear combination of the basis vectors define:

where and .

Now check that this satisfies the conditions in the definition of a inner product, and that the basis is ortho-normal with respect to this "inner product".

so <u1,u1> , our "new" inner product, is the dot product of the B-coordinate vectors in R^3:

<u1,u1> = (1,0,0).(1,0,0) = 1+0+0 = 1.

the ability to squeeze a scalar from two vectors in a positive-definite bilinear way depends on us assigning an array of scalars to any vector. the way we do this is, in effect, tantamount to choosing a basis.

because, for any given vector space, we can choose a basis in many different ways, we can also come up with many different inner products. you can actually go either way with this:

start with a given inner product, and adjust the basis to be orthonormal (Gramm-Schmidt), or start with a basis, and adjust the inner product to be orthonormal.

the key to this whole process, is understanding that for any field F, we can make a finite-dimensional vector space V "look like F^n" (that is we can find an isomorphism to F^n, or in numerical terms: create a change-of-basis matrix).

well, in F^n, everything is simple, because the standard basis is so "nice". so all CaptainBlack did here is make our basis behave itself through the (implicit) transformation:

uj <--->ej (here, to avoid confusion i am using uj for our "given" basis, and ej for the standard basis in F^n)

so to calculate the (new) inner product, we pull our linear combinations in B back to linear combinations in the standard basis F^n, perform the standard inner product (dot product), and then assign that value back to the new inner product: